\section{Further module theory}
The idea of this chapter is to put restrictions on arbitrary categories such that we will end up with categories of $MOD_R$ for some ring $R$.
\begin{Definition}[zero objects, pointed category]
An object $0$ of a category $\CC$ that is both initial and terminal is called a \textit{zero object} of $\CC$. If a category $\CC$ has a zero object it is called a pointed category.

\end{Definition}
Denote the unique morphisms $0\rightarrow C$ and $C\rightarrow 0$ as $0$, so $Mor_{\CC}(0,C)=\{0\}, Mor_{\CC}(C,0)=\{0\}$. Thus $0$ is unique up to unique isomorphisms (if it exists).

\begin{Example} We take a look at our zoo of categories.

\begin{itemize}
	\item $MOD_R$ has $0$ as a zero object.
	\item $GP$ has $1$ as a zero object.
	\item $RNG$ has $0$ as a zero object.
	\item  $RING$ has no zero object
	\item $TOP_0$ has singleton as its zero object.
	\item $FIELD$ has no zero object.
	\item $SET$ has no zero object.
	
\end{itemize}

\end{Example}
When $\CC$ has a zero object $0$, we can define the zero morphism between any two objects $C$ and $D$ to be the composite.

$$\xymatrix{
& 0 \ar[dr]^{\exists !} \\
C \ar[rr]_{0} \ar[ur]^{\exists ! } & & D} $$
This morphisms $0: C\rightarrow D$ is easily seen to be independent of choice of $0$ and has the property
that 
$$0\beta=0: B\rightarrow D \text{ and } \delta 0=0:C\rightarrow E$$

\begin{Definition}[Kernel, Cokernel]
Let $\gamma: C\rightarrow D$ be a morphism in a pointed category $\CC$.
Define a new category $KER(\gamma)$ with objects of the type $(B,\beta)$ and $(B',\beta')$ (where $B$ and $B'$ are objects of $\CC$ and $\beta: B\rightarrow C$ and $\beta': B'\rightarrow C$) and the morphisms $(B,\beta)\rightarrow (B',\beta')$ are given by the following diagramm
$$
\xymatrix{
B \ar[dd]_{\zeta} \ar[dr]_{\beta}\ar[drr]^{0}\\
& C \ar[r]_{\gamma}& D \\
B' \ar[ur]^{\beta'} \ar[urr]_{0}
} $$
A kernel of $\gamma$ is a pair $(Ker \gamma, K)$ and is a terminal object in $KER(\gamma)$ - if it exists.
\end{Definition}
Informally $\gamma\circ K=0 :  Ker(\gamma)\rightarrow D$ and it has to be terminal.
\begin{Example}
Consider in $AB, \pi: \ZZ \twoheadrightarrow \ZZ /2$. We can think of $Ker(\pi)$ as either $(2\ZZ, 2\ZZ \hookrightarrow \ZZ)$ or as ($\ZZ, \ZZ \stackrel{\cdot 2 }{\rightarrow } \ZZ)$. There exists a unique isomorphism:
$$\xymatrix{
\ZZ \ar[dr]^{\cd 2} \ar[dd]^{\cd 2}\\
& \ZZ \\
2\ZZ \ar@^{^{ (}->}[ur]
}
$$
\end{Example}
Dually we can define the cokernel.
\subsection{Additive and abelian categories}
\begin{Definition}[Preadditive categories]
Call a category $\CC$ preadditive if: 

\begin{itemize}[*]
	\item It is a pointed category and
	\item $Mor_{\CC}(M,N)$ is always an additive group, such that 
	\item for any morphisms $\lambda, \lambda': L\rightarrow M$ and $\mu,\mu': M\rightarrow N$
	 \begin{itemize}
	\item
	 \begin{itemize}
		 \item $\mu \circ (\lambda+\lambda') =\mu\circ \lambda + \mu \circ \lambda '$  or equivalently
		\item $\mu_*$ is a group homomorphisms, or equivalently
		\item $(\lambda+\lambda')^*=\lambda^*+\lambda'^*$
	 \end{itemize}
	
	\item
	 \begin{itemize}
		 \item $(\mu+\mu')\circ\lambda =\mu\circ \lambda + \mu' \circ \lambda $  or equivalently
		\item $\lambda_*$ is a group homomorphisms, or equivalently
		\item $(\mu+\mu')_*=\mu_*+\mu'_*$
	 \end{itemize}
	\item or equivalently $Mor_{\CC}(L,M)\oplus Mor_{\CC}(M,N)\rightarrow Mor_{\CC}(L,N)$, $(\lambda,\mu)\mapsto \mu \circ \lambda$ is a group homomorphism
	\end{itemize}
	
	
	 
\end{itemize}
Recall that $\lambda^*=Mor_{\CC}(\lambda,N): Mor_{\CC}(M,N)\rightarrow Mor_{\CC}(L,N)$ maps a morphisms $Mor_{\CC}(M,N)$ to a morphism in $Mor_{\CC}(L,N)$ by prefixing $\lambda$ and 
\mbox{$\mu_*=Mor_{\CC}(L,\mu):Mor_{\CC}(L,M)\rightarrow Mor_{\CC}(L,N)$} maps a morphism in $Mor_{\CC}(L,M)$ to a morphism in $ Mor_{\CC}(L,N)$ by postfixing $\mu$.

If $\CC$ is preadditive, then so is $\CC^{op}$.
\end{Definition}

\begin{Definition}[Mono, Epi]
Call a morphism

\begin{itemize}
	\item $\mu: X\rightarrow Y$ mono if $\forall L$ $0\rightarrow Mor_{\CC}(L,X)\stackrel{\mu_*}{\rightarrow}Mor_{\CC}(L,Y)$ is exact.
	\item $\epsilon: M\ra N$ epi if $\forall Z$ $0\ra Mor_{\CC}(N,Z)\stackrel{\epsilon^*}{\ra}Mor_{\CC}(M,Z)$ is exact.
	\item Exercise: Why not require that $Mor_{\CC}(N,Z)\stackrel{\epsilon^*}{\ra}Mor_{\CC}(M,Z)\ra 0$ be exact?
\end{itemize}
\end{Definition}
\begin{Example}

\begin{enumerate}
	\item Split epimorphisms and split monomorphisms.
	\item In the category of $R$-modules $MOD_R$ we have mono=injective and epi=surjective. 
	\item In the category of groups $GP$, we also have mono=injective and epi=surjective but the proof is not obvious (difficult exercise which needs generalization of above definitions since $GP$ is not preadditive).
	\item In the category of Hausdorff topological abelian groups and continuus homomorphisms, $\iota: \QQ\hookrightarrow \RR$ is epi (but obiously not surjective). It is also mono.
\end{enumerate}

\end{Example}

\begin{Lemma}
Let $\alpha: A\ra B$ be a morphism in a preadditive category $\CC$. Then

\begin{itemize}
	\item $\kappa: K\ra A$ is a kernel for $\alpha$ iff $\forall L$  in $\CC$
	$$0 \ra Mor_{\CC}(L,K)\stackrel{\kappa_*}{\ra}Mor(L,A)\stackrel{\alpha_*}{\ra}Mor(L,B) $$
is exact in $AB$. Thus for any choice of $Ker \alpha$,
$$Mor_{\CC}(L,Ker\alpha)=Ker(Mor_{\CC}(L,\alpha). $$
\item $\gamma: B\ra C$ is a cokernel for $\alpha$ iff $\ldots$
\end{itemize}
Hence, a morphism is mono iff its kernel is the zero object, and it is epi iff $\ldots$.
\end{Lemma}

A short exact sequence in a preadditive caregory is a sequence
$$0\ra A \stackrel{\alpha}{\ra}B\stackrel{\beta}{\ra}C \ra 0 $$ in which $(A,\alpha)$ is a kernel for $\beta$ and $(C,\beta)$ is a cokernel for $\alpha$.
In particular $\alpha$ is mono and $\beta$ is epi.
A subcategory $\CC'$ of a preadditive category $\CC$ is a preadditive category if

\begin{itemize}
	\item $Mor_{\CC '}(L,X)$ is a subgroup of $Mor_{\CC}(L,X)$ and
	\item $\CC'$ contains a zero object of $\CC$ (then $\CC'$ is a pointed category. 
\end{itemize}


\begin{Example} We give examples of subcategories of a preadditive category.
\begin{enumerate}
\item The zero object of a pointed category defines a preadditive category with one object.
\item A nonzero object $X$ of a preadditive category $\CC$ gives a minimal full preadditive subcategory $PREADD(X)$ of $\CC$; its objects are $0$ and $X$ and its morphisms are $End_{\CC}(X)$ together with the zero morphisms.
\end{enumerate}
\end{Example}
Next lecture: Additive categories and move on to abelian categories.

\begin{Example}[Don't understand this one. Did I get it wrong?]
This might be a collection of results?
\textcolor[rgb]{0,0,1}{Yes, it is a collection of results. Should form a table instead...}
\begin{itemize}
\item pointed category: there exists a zero object.
	\item preadditive: $Mor_{\CC}(M,N)$ is a additive group. $\mu_*$ is a group homomorphism: preadditive.
	\item additive: Exists $M'\oplus M''$ \textcolor[rgb]{1,0,0}{(What does that mean?)}\textcolor[rgb]{0,0,1}{This just means that for every pair of objects $M$ and $M'$, the direct sum of $M$ and $M'$ exists}
	\item abelian
	\begin{itemize}
		\item Every morphism has a kernel and a cokernel
		\item every monomorphism is a kernel
		\item every epi is a cokernel
	\end{itemize}
	
\end{itemize}


\end{Example}

\subsection{Additive category}
Recall Tutorial 6 Exercise 7: 
\begin{Definition}[direct sum]
A direct sum $M=M'\oplus M''$ is characterized by morphisms $\sigma':M'\rightarrow M$, $\pi':M\rightarrow M'$ $\sigma'':M''\rightarrow M$, $\pi'': M \rightarrow M''$ such that $\pi'\sigma'=id_{M'}$ and $\pi''\sigma''=id_{M''}$ and $id_M=\sigma'\pi'+\sigma''\pi''$.
These condition make sense in any preadditive category. When they hold, call $M$ the direct sum of $M'$ and $M''$.
\end{Definition}
\begin{Definition}[additive category]
Let $\CC$ be a preadditive category. We call $\CC$ additive if all finite direct sums of objects in $\CC$ exist.
\end{Definition}

\begin{Example}
The full subcategory $PREADD(\ZZ)$ of $AB$ is an example of a preadditive category in which the direct sum is missing. Its only objects are $0$ and $\ZZ$.
\end{Example}
\begin{Definition}[split sequence]
Assume the category $\CC$ is preadditive. Call an exact sequence
$$0\rightarrow M' \stackrel{\sigma'}{\rightarrow} M \stackrel{\pi''}{\rightarrow}M''\rightarrow 0 $$
split iff there exists $\sigma'':M''\rightarrow M$ with $\pi''\sigma''=id_{M''}.$ 
\end{Definition}
 The existence of a $\pi':M\rightarrow M''$ such that $\pi'\sigma'=id_{M'}$ provides an equivalent definition of a split sequence.

\begin{Lemma}
Let $M, M', M''$ be objects in a preadditive category $\CC$. The following are equivalent:

\begin{itemize}
	\item $M$ is isomorphic to the direct sum of $M'\oplus M''$.
	\item $M$ is the product of $M'$ and $M''$.
	\item $M$ is the coproduct of $M'$ and $M''$
	\item There exists a split exact sequence
	$$0\rightarrow M' \stackrel{\sigma'}{\rightarrow} M \stackrel{\pi''}{\rightarrow}M''\rightarrow 0 $$
\end{itemize}
\end{Lemma}
We call a preadditive category $\CC$ an additive category if every pair of objects $M'$ and $M''$ has a direct sum. An additive subcategory $\CC'$ is a preadditive subcategory $\CC'$ of $\CC$ such that $\CC'$ is additive.
\begin{Lemma}
The categories $PROJ_R$, $FREE_R, M_R, J_R, P_R$ are additive subcategories of $MOD_R$. If $S$ is a a subring of $R$, then $MOD_R$ is an additive subcategory of $MOD_S$.

In the category of abelian groups $AB$, $PREADD(\ZZ)$ is a preadditive subcategory of $AB$, but is not an additive subcategory.
\end{Lemma}
\begin{Definition}[Additive functor]
Let $\CC$ and $\DD$ be additive categories. An additive functor $F:\CC\rightarrow \DD$ is one that preserves this additive structure, i.e. $F:Mor_{\CC}(C',C)\rightarrow Mor_{\DD}(FC',FC)$ is a group homomorphisms $\forall C',C$ in $\CC$ $F(\alpha+\beta)=F(\alpha)+F(\beta)$
\end{Definition}

\begin{Lemma}
An additive functor $F:\CC\rightarrow \DD$ preserves zero morphisms, zero objects and directs sums.
\end{Lemma}
\begin{Example}
In an additive category $\CC$, the known morphism functors provide examples of additive functors.
\begin{itemize}
\item  For a fixed $C$ in $\CC$, $Mor(C,-):\CC\rightarrow \CC^{op}$ is a covariant additive functor.
\item For fixed $C$ in $\CC$, $Mor(-,C):\CC^{op}\rightarrow \CC$ is a contravariant additive functor.
\item If $\CC'$ is an additive subcategory of $\CC$, then the inclusion functor is additive. Furthermore if $F:\CC\rightarrow \DD$ is additive, then so is the restriction of $F_{|\CC'}$ to $\CC'$.
\end{itemize}
\end{Example}
\subsection{Abelian Category}

\begin{Definition}[normal morphism, conormal morphism]
A monomorphism is called normal if it is the kernel of some morphism. 

An epimorphism is called conormal if it is the cokernel of some morphism.
\end{Definition}
\begin{Definition}[abelian category]
\label{abelianCategory}
An additive category $\CC$ is an abelian category if the following three conditions hold.

\begin{enumerate}[a)]
	\item For every morphism $\lambda: L\rightarrow M$ in $\CC$, $\CC$ contains a kernel $(Ker \lambda, K)$ and a cokernel $(Coker\lambda, X)$ of $\lambda$. So  $K: Ker \lambda \rightarrow L$ and $X: M\rightarrow Coker \lambda$ lie in $\CC$. 
	\item Every monomorphism is normal, i.e. if $\KK:K\rightarrow L$ is mono then $(K,\KK)$ is the kernel of some morphism $\lambda: L\rightarrow M$ in $\CC$. 
	\item Every epimorphisms is conormal, i.e. if $X: M\rightarrow C$ is epi, then $(C,X)$ is the cokernel of some $\lambda: L\rightarrow M$ of $\CC$.
	
	\end{enumerate}
\end{Definition}
If $R$ is a ring $MOD_R$ is an abelian category \textcolor[rgb]{1,0,0}{(is this correct?)} \textcolor[rgb]{0,0,1}{Yes, the motivation for abelian category comes from the category $MOD_R$}. The following Lemma states conditions such that a full subcategory of $MOD_R$ is abelian.
\begin{Lemma}
Let $R$ be a ring and $\CC$ a full subcategory of $MOD_R$. Then $\CC$ is abelian iff $\forall$ homomorphisms $\lambda:L\rightarrow M$ of $R$-mod in $\CC$, the $R$-modules $Ker \lambda,~ Cokernel \lambda$ are in $\CC$. 
\end{Lemma}
\begin{proof}We proof the conditions in Definition \ref{abelianCategory}.
\begin{itemize} 
\item[a)]
The $R$-modules $Ker \lambda$, $cokernel \lambda$ satisfy the universal property with regard to all $R$-modules and $R$-homomorphisms in $MOD_R$. In paricular they satisfy the universal property with regard to those in $\CC$. Thus condition a) holds.
\item[b)]
Consider a monomorphism $\KK: K\rightarrow L$, i.e. $\forall~ W$ in $\CC$ the sequence
$$0\rightarrow Mor_{\CC}(W,K) \stackrel{\KK}{\rightarrow} Mor_{\CC}(W,L)$$ is exact in $AB$.
We claim that $(K,\KK)$ is the kernel of $\pi: L\twoheadrightarrow L/Im(\KK)$.

We first show that $\KK$ is injective. Put $W=Ker_{\CC}(\KK)$ gives $Mor_{Mod_R}(Ker \KK,K)=Mod_{\CC}(
Ker_{\CC}\KK,K)=\{0\}$.
Since in $MOD_R$, $Ker \KK \hookrightarrow K$, we deduce $Ker \KK=0$
Thus $\KK$ is injective and $Im \KK\cong K$.
Then $Coker \KK\cong L/Im(\KK)$. Since $\KK$ is in $\CC$, so Cokernel $\KK$ is in $\CC$. Then $(K,\KK)$ is kernel of $\pi: L\twoheadrightarrow L/Im(\KK)$. Since $\CC$ is a full subcategory, $\pi$ is in $\CC$. So its kernel $(K,\KK)$ is in $\CC$.
\textcolor[rgb]{1,0,0}{I really do not understand this proof yet.}
\item [c)] similarily.
\end{itemize}
\end{proof}
\begin{Example}
The category of Hausdorff topological abelian groups is an example of a full additive subcategory of $AB$ that is not abelian.
\end{Example}

\subsection{Embedding Theorems}
This chapter states theorems of embedding one category into another.
\begin{Definition}[faithful functor, full functor]
Call a functor $F:\CC\rightarrow \DD$ 
\begin{itemize}
\item faithful if the induced set function $Mor_{\CC}(C',C)\rightarrow Mor_{\DD}(FC',FC)$ is always injective. (e.g. inclusion of a subcategory, most forgetful functors) and 
\item full if the induced set function $Mor_{\CC}(C',C)\rightarrow Mor_{\DD}(FC',FC)$ is always surjective (e.g. the inclusion of a full subcategory).
\end{itemize}
\end{Definition}
\begin{Definition}[concrete category]
A category is called concrete if there is a faithful functor to the category $SET$.
\end{Definition}
\begin{Lemma}
 If $\CC$ is a small additive category, then there exists a contravariant embedding (=faithful functor) in the functor category $\left[\CC,AB\right]$ via \mbox{$C\mapsto Mor_{\CC}(C,-)$}.

Since $\CC$ is additive $\CC^{op}$ is additive and one gets a covariant embedding of $\CC$ in the additive category $\left[\CC^{op},AB\right]$ by $C\rightarrow Mor_{\CC}(-,C)$.

\end{Lemma}
\begin{Theorem}[Lubbin-Henn-Freyd Representation Theorem - also Mitchell, 1960s]
Any small abelian category can be embedded in $AB$. 
\end{Theorem}
\begin{Theorem}
Given a small abelian category $\CC$, there exists a ring $R$ and a covariant full, faithful functor $F: \CC\rightarrow MOD_R$, i.e. $Mor_{\CC}(C',C)\stackrel{\cong}{\rightarrow}Mor_{Mod_R}(FC',FC)$.

This functor is exact in that it presevers exact sequences.
\end{Theorem}

\subsection{Tensor Product}
Definitions can be found in MacLane and Bichoff's book 'Algebra' in Chapter $IX$.
Tensor products arise from bilinear maps.
\begin{Example}
Field $F$, $F^n\times F^n\rightarrow F$. Scalar dot product $(u,v)\mapsto u\cd v$
\begin{itemize}
\item $(u+u')\cd v=u\cd v + u' \cd v$.
\item $u\cd (v+v')=u\cd v + u \cd v'$.
\item $(\alpha u)\cd v= u\cd (\alpha v)=\alpha (u\cd v)$.
\end{itemize}

\end{Example}
The Tensor product $F^n\otimes_F F^n$ of a vector space $F^n$ with itself has the universal property that \mbox{$F^n\times F^n \rightarrow F^n\otimes F^n$} is initial among all biadditive balanced functions from $F^n\times F^n$. \textcolor[rgb]{1,0,0}{(What are balanced functions?)} \textcolor[rgb]{0,0,1}{According to prof, they are functions that satisfies the third condition above}

Why do we study $- \otimes -$?
We use its adjunction as motivation. Then, we derive its biadditive and balanced properties. 


We use Fn(X,Y) as abbreviation for $Mor_{SET}(X,Y)$.
\begin{Lemma}
There is a bijection \textcolor[rgb]{1,0,0}{(or what do we have here? A functor?)} \textcolor[rgb]{0,0,1}{Bijection is correct. Though this shouldn't be a lemma. It is just what motivates the construction of tensor product.}
$$Fn(X\times Y, Z)\longleftrightarrow Fn(X,Fn(Y,Z))$$ by 

$$\left[(x,y)\mapsto f(x,y)\right]\mapsto \left[x\mapsto\left[y\mapsto f(x,y)\right]\right] \text{ and}$$

$$\left[(x,y)\mapsto \alpha_x(y)\right]\mapstoreverse \left[x\mapsto\alpha_x\right].$$

The functor $-\times Y$ is left adjoint to $Fn(X,-)$.

$$Fn(M\times F,B)\longleftrightarrow Fn(M,Fn(F,B)) \longleftrightarrow Fn(F,Fn(M,B)).$$
\end{Lemma}
\begin{Example} We give several examples for the above Lemma.

\begin{itemize}
	\item 

Field $F$, vector space $F^n$, ring $M_n (F)$ of $n\times n$ matrices.
There exists adjunction
$$Fn(M_n(F)\times F^n, F^n) \longleftrightarrow Fn(M_n(F),Fn(F^n,F^n))$$
On the LHS, we have $(P,\textbf{v}) \mapsto P\textbf{v}$ and on the RHS, we have $P \mapsto [\textbf{x} \mapsto P\textbf{x}]$ given by 
$$m:M_n(F)\times F^n\rightarrow F^n  \text{ and }\hat{m}: M_n(F)\rightarrow End(F^n)$$ which are ring isomorphism.
\textcolor[rgb]{1,0,0}{(needs some fixing here)} \textcolor[rgb]{0,0,1}{Noted. Fixed some stuff}

\item Consider the category $LGC$ of all logical statements. Say there exists a morphism from statement $X$ to statement $Y$ precisely when $X$ implies $Y$, written $X\Rightarrow Y$. 
$Mor_{LGC}(X\wedge Y,Z)\mapstoreverse MOR_{LGC}(X,Mor_{LGC}(Y,Z))$.
\item 
However $CTS(\RR x\times \RR y, \RR) \longleftrightarrow Fn(\RR x, \RR y, \RR)$. \textcolor[rgb]{1,0,0}{(There has to be something wrong in here!)} \textcolor[rgb]{0,0,1}{There is! It should be that there is no bijection between the two sets} E.g. In RHS, 

\[
  \alpha_x(y) = \left\{
  \begin{array}{l l}
    \frac{2xy}{x^2+y^2}  & \quad \text{if $x\neq$ 0}\\
    0 & \quad \text{if x=0}\\
  \end{array} \right.
\]

\[
  f(x,y) = \left\{
  \begin{array}{l l}
    \frac{2xy}{x^2+y^2} & \quad \text{if $(x,y)\neq (0,0)$}\\
    0 & \quad \text{if (x,y)=(0,0)}\\
  \end{array} \right.
\]

Thus $\theta \neq 0$ \textcolor[rgb]{1,0,0}{(what is that supposed to mean?)} gives a discontinuity at $0$ so $f \not\in LHS$.
\item 
$TOP$, the challenge is to find a topology for $X\times Y$ and for $CTS(Y,Z)$ that allows $CTS(X\times Y, Z)\leftrightarrow CTS(X,CTS(Y,Z))$.

Partial solution: Product topology on $X\times Y$ and the compact open topology on $CTS(Y,Z)$ but this works only for a reasonable space.
\item In $TOP_0$, replace $X\times Y$ with $X\wedge Y=(X\times Y)$ and the smash product. \textcolor[rgb]{1,0,0}{(Whatever that is!)} \textcolor[rgb]{0,0,1}{I have no clue what is it. I don't know topology}
\item In $AB$ (Whitney 1938), $Hom(H,K)$ is an abelian group by adding values \textcolor[rgb]{1,0,0}{(what is adding values?)} \textcolor[rgb]{0,0,1}{i.e. the homomorphism form an abelian group by adding two functions in the usual manner}. So we can try for 
\textcolor[rgb]{0,0,1}{Here we are trying to aim for a result similar to the above discussion. So we are trying to figure out what is the correct thing to replace the ``?''. So, we do some computations.}
$Hom(G?H,K)\leftrightarrow Hom(G,Hom(H,K))$


Example: Let $G=H=K=\ZZ/2$. Then $Hom(\ZZ/2,\ZZ/2)\cong \ZZ/2$.
So $Hom(G,Hom(H,K))\cong \ZZ/2$

But, $Hom(\ZZ/2\times \ZZ/2)\cong \ZZ/2 \times \ZZ/2$, so the ``?'' is not direct product. 

Since $Hom(\ZZ/2,\ZZ/2)\cong \ZZ/2$, this suggests that $\ZZ/2?\ZZ/2\cong \ZZ/2$.
\textcolor[rgb]{1,0,0}{CAN SOMEONE TAKE A LOOK AT THIS EXAMPLE? I THINK THERE ARE SEVERAL MISTAKES IN HERE!}  \textcolor[rgb]{0,0,1}{I've done some modifications}
\end{itemize}

\end{Example}

Construction: For $L_R  \in MOD_R$, ${}_R M \in {}_R MOD$, we want to construct $L\otimes _R M \in AB$, such that 
$\forall N\in AB$ there exists a natural isomorphisms
\mbox{$\eta: Mor_{AB}(L\otimes_R M, N) \stackrel{\cong}{\longrightarrow} MOR_{MOD_R}(L_R,Mor_{AB}({}_R M,N))$}

(Recall that \mbox{$Mor_{AB}({}_R M,N) \in MOD_R$: $(\gamma r) m=\gamma (rm)$})

$$\xymatrix{
Mor_{MOD_R}(L_R,MOD_{AB}({}_R M, N)) \ar[rdd]_{\theta} \ar@{>->}[r] & Fn(\Upsilon L, Fn(\Upsilon M, \Upsilon N)) \ar@{<->}[d]\\
& Fn(\Upsilon L \times \Upsilon M, \Upsilon N) \ar@{<->}[d]^{\cong~by~Fr_{\ZZ}} \\
& Mor_{AB}(Fr_{\ZZ}(\Upsilon L\times \Upsilon M, N)
}$$

Hence $\forall N\in AB$, $\theta\eta: Mor_{AB}(L\otimes_RM, N) \hookrightarrow  Mor_{AB}(Fr_{\ZZ}(\Upsilon L\times \Upsilon M),N)$.
How to describe the natural monomorphism  $\theta\eta$?
\begin{Lemma}[Yoneda's Lemma]
Every natural transformation $\chi$ from $Mor(H,-)$ to $Mor(K,-)$ has the form $\epsilon^*$ for $\epsilon=I(id_H)$
$$
\xymatrix{
K \ar[r]{\epsilon} \ar[dr]{\epsilon^*_{\alpha}}& H \ar[d]{\alpha}\\
& L
}$$


Here for $H=L\otimes_r M$ let $\epsilon=0\eta(id_{L\otimes_R M}): Fr_{\ZZ}(\Upsilon L \times \Upsilon M)\rightarrow L\otimes _R M$ be a homomorphism of abelian groups.

By Yoneda's Lemma, $0\eta=\epsilon^*$ since $\epsilon^*$ is mono. By definition, $\epsilon$ is epi in $AB$, i.e. $\epsilon$ is surjective homomorphism $Fr_{\ZZ}(\Upsilon L \times \Upsilon M)\rightarrow L\otimes_R M$.
because  $L\otimes_R M=Fr_{\ZZ}(\Upsilon L \times \Upsilon M)/r$ for some equivalence relation that we have to find.
Write the equivalence class of $(l,m)$ as $l\otimes m$.

A typical element of $Fr_{\ZZ}(\Upsilon L \times \Upsilon M)$ has the form $\Sigma' k_i(l_i,m_i),~k_i\in \ZZ$ so a typical element of $ L \otimes_R  M$ has the form $\Sigma' k_i(l_i\otimes m_i),~k_i\in \ZZ$
\end{Lemma}
After Yoneda consider $\eta(id_{L\otimes_R M})$. 
$\eta(id)(l)=[m\mapsto id(l\otimes m)=l\otimes m] \in Mor_{AB}({}_R M, L\otimes_R M)$, i.e. $l\otimes m=\eta(id)(l))(m)$
and 

\begin{itemize}
	\item for each $l$, $\eta(id)(l)$ preserves addition. Also $\eta(id)\in Mor_{Mod_R}(L_R,Mod_{AB}({}_R M, L\otimes_R M))$
	\item $\eta(id)$ preservers addition
	\item $\eta(id)$ preservers right multiplication by $R$
	
	\begin{itemize}
		\item $l\otimes (m_1+m_2)=\eta(id)(l)(m_1+m_2)=\eta(id)(l)(m_1)+ \eta(id)(l)(m_2)=l\otimes m_1+l\otimes m_2$
		\item $(l_1+l_2)\otimes m=\ldots = l_1\otimes m + l_2 \otimes m$
		\item $(lr)\otimes m=\ldots=l\otimes (rm)$
	\end{itemize}
	
\end{itemize}

\begin{Theorem}
Let $M$ be a left $R$-module. A functor $\otimes_R M: Mod_R\rightarrow AB$ is left adjoint to $Mor_{AB}({}_R M, -)$ iff forall right $R-$modules $L_R$ we have
\mbox{$L\otimes_R M= \{\Sigma'k_i(l_i\otimes m_i) |k_i\in \ZZ a,b,c\}$.}
Moreover (check!) $\alpha:L'\rightarrow L$ gives
$\alpha\otimes_R M: L'\otimes_RM \rightarrow L\otimes_R M$
$\Sigma'k_i(l'\otimes m_i)\mapsto \Sigma'k_i(\alpha(l')\otimes m_i)$
\end{Theorem}

Basic properties of $\otimes$.
We now extend the definition to allow for a right $s$ action and a left $\QQ$ action.

\begin{Theorem}
 If also $M\in {}_R BIMOD_S$ and $N\in MOD_S$, then
$\eta$: $Mor_{...}(L\otimes_R M_S,N_S)\stackrel{\cong}{\longrightarrow}Mor_{MOD_R}(L_R,Mor_{MOD_S}({}_R M_S,N_S)$. If further $L\in {}_{\QQ} BIMOD_R$ and $N\in {}_{\QQ}BIMOD_S$, then
: $Mor_{{}_{\QQ}BIMOD_S}({}_Q L_R \otimes_R M_S,{}_{\QQ} N _S)\stackrel{\cong}{\longrightarrow}Mor_{{}_{\QQ}BIMOD_R}({}_{\QQ}L_R, MOR_{MOD_S}({}_R M _S, {}_{\QQ} N _S).$

Here we use $q(lr\otimes m)s=ql\otimes rms$ and $((\lambda (qlr))(m))s=q(\lambda (l))(rms).$
\end{Theorem}
\begin{Example}
\begin{itemize}
\item T 4.8 $Q=S=\ZZ$
\item When $R$ is commutative we take $Q=R=S$. The equations shows that the biadditive maps are actually bilinear.
Let $R,S$ be arbitrary rings with $M\in {}_R BIMOD _S$. As above forall right $R$-modules $L$, $L\otimes_RM$ is a right $S$-module via
($\otimes m) s =l\otimes (ms)$.
For all right $R$-module homomorphisms $\alpha: L'\rightarrow L$, the induced homomorphisms $\alpha\otimes id_M: L'\otimes_R M \rightarrow L\otimes _R M$ defined by 
($\alpha\otimes id_,)(l'\otimes m)=\alpha(l')\otimes m$ is a homomorphism of right $S$-module. Sometimes write a mapping from the objects of $MOD_R$ to the objects of $MOD_S$ and a collection of mappings.
\end{itemize}
\end{Example}



\begin{Lemma}
$-\otimes_R M_S$ is an additive functor from $MOD_R$ to $MOD_S$.
\end{Lemma}
\begin{proof}
$-\otimes_R M: Hom_R(L',L)\rightarrow Hom_S(L'\otimes_RM, L\otimes_RM)$ one for each pair of $R$-modules $L$ and $L'$.
Is this a functor? Clearly $id_L\otimes id_M=id_{L\otimes M}$ and $\beta\alpha\otimes id_M=(\beta\otimes id_M)(\alpha\otimes id_M).$
and so $-\otimes_R M$ defines a functor from $MOD_R$ to $MOD_S$. Moreover $-\otimes_R M$ is additive, because if $\alpha,\alpha':L'\rightarrow L$ are right $R$-modulehomomorphisms, then $(\alpha+
\alpha')\otimes id_M=\alpha\otimes id_M+\alpha'\otimes id_M$.
\end{proof}

\begin{Lemma}
Let $L=\oplus_{\lambda\in \Lambda}L_\lambda$ be the direct sum of right $R$-modules where $\Lambda$ is any ordered set.
$L\otimes_R M\cong \oplus_{\lambda\in \Lambda}(L_{\lambda}\otimes_R M)$.
\end{Lemma}
\begin{Proof}
Clear from previous Lemma for finite $\Lambda$. Exercise for infinite $\Lambda$.
\end{Proof}

\begin{Proposition}

\begin{itemize}
	\item The functor $-\otimes_R R$ is naturally isomorphic to the identity functor on $MOD_R$.
	\item The functor $R\otimes_R -$ is natrually isomorphic to the identity functor on ${}_R MOD$.
\end{itemize}

\end{Proposition}

\begin{Proof}
For all right $R$-modules $L$, let $\beta_L: L\otimes_R R \rightarrow L$ be the $R$-module homomorphism given by $l\otimes r\mapsto lr$.
If $\alpha: L'\rightarrow L$ is a homomorphisms of right $R$-modules then
$$
\xymatrix{
L'\otimes R \ar[r]^{\beta_{L'}} \ar[d]_{\alpha\otimes id_M}& L' \ar[d]^{\alpha}\\
L\otimes R \ar[r]_{\beta_L}& L
}$$
Thus $\beta$ is a natural transformation from $-\otimes_R R$ to $ID_{MOD_R}$
Observe that it has the inverse $L\rightarrow L\otimes_R R, l\mapsto l\otimes 1$
\end{Proof}

\subsection{Localization of $R$-modules}
Grell 1927, Ore 1933, Krull 1938

Aim: to construct from a ring $R$ and a suitable subset $\Sigma$ of $R$ a ring $\Sigma^{-1}R=R_{\Sigma}=R[\Sigma^{-1}])$ in which elements of $\Sigma$ have inverses. Later do the same for $R$-modules, enables us to focus on cerain parts of the module and ignore others.
\begin{Example} 
Compare the two abelian groups of order 12:
$\ZZ/4\ZZ\oplus \ZZ /3\ZZ$ and $\ZZ /2\ZZ\oplus \ZZ /2\ZZ  \oplus \ZZ /3\ZZ$.
One way to distinguish them is to take \mbox{$-\otimes_{\ZZ} \ZZ /2\ZZ$}.

$3$ is invertible in $\ZZ /2\ZZ$ but $3$ is zero in $\ZZ /3\ZZ$ so in
$\ZZ /3\ZZ\oplus \ZZ /2 \ZZ$, $1\otimes 1=1\otimes 3\cd 3=3\otimes 3=0\otimes3=0\otimes 0$. So 
$\ZZ /3\ZZ \otimes_{\ZZ} \ZZ /2\ZZ=0$.

Thus $-\otimes_{\ZZ}\ZZ /2\ZZ$ allows us to focus on the 2-primary part where
$$\ZZ /2\ZZ \otimes_{\ZZ} \ZZ /2\ZZ= \ZZ /2\ZZ\not \cong \ZZ /2\ZZ\oplus \ZZ /2\ZZ$$
$$=(\ZZ /2\ZZ\otimes_{\ZZ} \ZZ /2\ZZ) \oplus (\ZZ /2\ZZ\otimes_{\ZZ}\ZZ /2\ZZ)=(\ZZ /2\ZZ\oplus\ZZ /2\ZZ)\otimes_{\ZZ}\ZZ /2\ZZ.$$
However, $-\otimes_{\ZZ} \ZZ /2\ZZ$ is too drastic, since it fails to distinguish between $\ZZ /2\ZZ\oplus \ZZ /8\ZZ$ and $\ZZ /4\ZZ\oplus \ZZ /4\ZZ$.
Hence, we need to take $-\otimes_{\ZZ} O$ with minimal/intial $O$ in which 3 becomes invertible. (Actually, to focus on the 2-primary part, we want all primes $\neq 2$ to become invertible). This is like $\ZZ\hookrightarrow \QQ$ where we make all primes invertible.

\end{Example}

From now on we consider all rings to be commutative.
\begin{Definition}[multiplicative set]
For any ring $R$ a multiplicative subset $\Sigma \subset R$ is a submonoid of $(R,\cd)$.
\end{Definition}
Given an $R$-module $M$  we define $(m,\sigma)\mapsto m/\sigma$

\begin{Definition}[localization]
We define $\Sigma^{-1} M:=(M\times \Sigma) / \sim$ where $\sim$ is defined by

$(m_1,\sigma_1)\sim (m_2,\sigma_2) \Leftrightarrow \exists \nu\in \Sigma$ with $m_1\sigma_2\nu=m_2\sigma_1\nu$


\end{Definition}
\begin{Lemma}
$\Sigma^{-1}M$ is an abelian group via
$m/\sigma + m'/\sigma'=(m\sigma'+m'\sigma)/\sigma\sigma'$.
\end{Lemma}
\begin{proof}
Suppose $m_1\sigma_2 \nu =m_2\sigma_1\nu$. 
Show $(m_1\sigma'+m'\sigma_1,\sigma_1\sigma')\sim (m_2\sigma'+m'\sigma_2,\sigma_2\sigma')$, i.e. $\exists \mu$ with
$(m_1\sigma'+m'\sigma_1)\sigma_2\sigma'\mu=(m_2\sigma'+m'\sigma_2)\sigma_1\sigma'\mu$
i.e.
$m_1 (\sigma')^2 \sigma_2\mu+m' \sigma_1\sigma_2\sigma'\mu=m_2\sigma_1(\sigma')^2+m'\sigma_1\sigma_2\sigma'_\mu$

i.e. $m_1\sigma_2\sigma'^2\mu=m_2\sigma_1\sigma'^2\mu$ (because put $\mu=\nu$).
\end{proof}
Take $M=R_R$. Put $(r/\sigma)\cd (r'/\sigma')=rr'/\sigma\sigma'$. Then $\Sigma^{-1}R$ is a commutative ring with $1_{\Sigma^{-1}R}=1/1$.
$(m/\sigma)\cd(s/\tau)=ms/\sigma\tau$.
$\Sigma^{-1}M\times \Sigma^{-1}R \rightarrow \Sigma^{-1}M$
makes $\Sigma^{-1}M$ a $\Sigma^{-1}R$-module.

Define $j: R\rightarrow \Sigma^{-1}R, r\mapsto r/1$.

This process is called localization at $\Sigma$ and the ring $\Sigma^{-1} R$ is called the ring of fractions (of $R$ with respect to $Sigma$).

\begin{Example}

\begin{itemize}
	\item If $\Sigma\subset R^*$ then the elements in $\Sigma$ already have inverses in $R$, so $\Sigma^{-1}R\cong R$.
	\item If $0\in \Sigma$ then $\Sigma^{-1}R=0$.
	\item Let $R$ be a commutative domain, and put $\Sigma=R -\{0\}$. Then $\Sigma^{-1}R$ is the field of fractions of $R$.
	\item Let $\pp$ be a proper ideal of $R$. Then $\Sigma=R-\pp$ is multiplicative $\Leftrightarrow \pp$ is a prime ideal $\Leftrightarrow ( xy\in \pp \rightarrow x\in \pp  \vee  y\in \pp$).
	
	Recall $m$ maximal $\Leftrightarrow$ $R/m$ is a field $\Rightarrow R/m$ is a domain $\Leftrightarrow $m is prime.
	When $\Sigma=R-\pp$, we usually write $\Sigma^{-1}R$ as $R_{\pp}$ and speak of the localization of $R$ at $\pp$
	
	When $R$ is a domain and we define  $K:=R_{(0)}$ then for all prime ideals $\pp$ of $R$, the ring $R_{\pp}$ is a subring of $K$. 
	\item In a polynomial ring $R=S[t]$, taking $\Sigma=<t>$ (monoid of powers of $t$), $\Sigma^{-1}R=S[t,t^{-1}]=S[t,u]/(tu-1,ut-1)$is the Laurent polynomial ring.
	
	When $S$ is a domain, this lies in the field of rational polynomials $R_{(0)}=S(t)$, the field of fractions of $R$.
\end{itemize}


\end{Example}

\begin{Definition}[inverting ring homomorphism]
We say that a ring homomorphism $f: R\rightarrow S$ with $\Sigma\subset R$ of commutative rings is $\Sigma$-inverting if $f(\Sigma)\subset S^*$.
\end{Definition}

\begin{Theorem}
$\Sigma^{-1}R$ is the universal ring in which the elements of $\Sigma$ become invertible. More precisely, there exists a canonical homomorphism \mbox{$j: R\rightarrow \Sigma^{-1}R$} that is $\Sigma$-inverting and if $f$ is any $\Sigma$-inverting ring homomorphism, then
there is a unique induced ring homorphism $\barf: \Sigma^{-1}R\rightarrow S$ with $\barf j=f$.
$$
\xymatrix{
R \ar[r]^{j} \ar[dr]_{f} & \Sigma^{-1}R \ar[d]^{\exists ! \barf} \\
& S
}
$$
That is $j: R\rightarrow \Sigma^{-1}R$ is initial in the category of $\Sigma$-inverting ring homorphisms from $R$.
\end{Theorem}

\begin{Lemma}
When $R$ is a domain and $0 \not \in \Sigma$, then $j:R\rightarrow \Sigma^{-1}R$ is an embedding. 
\end{Lemma}

\begin{Proof}
Choose $S=R_{(0)}$. Here $f$ is injective, so $j$ must be too. 
\end{Proof}

\begin{Lemma}
\textcolor[rgb]{1,0,0}{What does this Lemma say? Have to add that information here...}
\begin{itemize}
	\item There exists a functor $MOD_R\rightarrow MOD_{\Sigma^{-1}R}$, $M\mapsto \Sigma^{-1}M$, $\left[\alpha: M'\mapsto M\right]\mapsto \left[\Sigma^{-1}\alpha: \Sigma^{-1}M'\rightarrow \Sigma^{-1}M\right], m'/\sigma\mapsto \alpha(m')/\sigma$. Check that well-defined preserves addition and scalar multiplication.
	\item Moreover by restriction of scalars $j^{\#}$, there exists $R$-homomorphism \mbox{$\Phi_M: M\rightarrow j^{\#}(\Sigma^{-1} M)$} $, m\mapsto m/1$ that defines a natural transformation \mbox{$\Phi: Id_{Mod_{R}}\Rightarrow j^{\#}\Sigma^{-1}: MOD_R\rightarrow MOD_R$}.
\end{itemize}

\end{Lemma}

\begin{Lemma}
$Ker (\Phi_M)=\{m\in M | \exists \nu \in \Sigma \text{ with } m\nu=0  \}.$ 
\end{Lemma}

\begin{Example}
For $R=\ZZ$ and $\Sigma=\ZZ-2\ZZ=\ZZ-(2)$ (complement of a prime ideal).

$Ker [\Phi_M: M\rightarrow M_{(2)}]\stackrel{4.16}{=}\{m\in M | \exists \text{  odd } \mu\in \ZZ \text{ with }  mv=0\}$
on a finitely generated group $M$, $2$-localozation $M\rightarrow M_{(2)}$ kills all odd primary torsion of $M$ (like $-\otimes_{\ZZ} \ZZ/2$) but (unlike - $-\otimes_{\ZZ} \ZZ/2$ is injective on the other summand (the torsion free part) by using 4.16, so we don't loose any information. 
Hence, it discriminates between $\ZZ/4\oplus \ZZ/4\oplus \ZZ/3$ and $\ZZ/8 \oplus\ZZ/2 \oplus \ZZ/3$.
(we just have the 2-primary left and get rid of the 3-primary part that we don't want)
\end{Example}

\begin{Proposition}
Suppose that $\alpha: M\rightarrow N$ is a $R$-homomorphism
Then

\begin{itemize}
	\item $Ker(\Sigma^{-1}\alpha)=\Sigma^{-1}Ker(\alpha) \subset \Sigma{^-1}M$,
	\item $Im(\Sigma^{-1}\alpha)=\Sigma^{-1}Im(\alpha)\subset \Sigma^{-1}N$ and
	\item $Coker(\Sigma^{-1}\alpha)=\Sigma^{-1}Coker(\alpha)$.
\end{itemize}
\end{Proposition}
\begin{Proof}$\Sigma^{-1}Ker\alpha:=\{m/\sigma | \alpha(m)=0\}$.
\begin{itemize}
\item[$\supseteq$]
If $m/\sigma \in \Sigma^{-1}Ker \alpha$, i.e. $\alpha(m)=0$, then $(\Sigma^{-1}\alpha)(m/\sigma)=\alpha(m)/\sigma=0/\sigma=0$.
Thus $\Sigma^{-1}(Ker \alpha) \subset Ker(\Sigma^{-1}\alpha)$.
\item[$\subseteq$]
$Ker(\Sigma^{-1}\alpha):=\{m/\sigma | \alpha(m)/\sigma=0\}=\{m/\sigma | \exists \nu\in\Sigma\text{ with }\alpha(m)\nu=0\} $
=$\{m/\sigma \exists v\in \Sigma$ with $\alpha(m\nu)=0\}=\{m\nu/\sigma\nu |\alpha(m\nu)=0\}\subset\Sigma^{-1}Ker(\alpha)$.
\end{itemize}
Other parts are similar, e.g. use $n/\sigma+\alpha(m)/\sigma'=(n\sigma'+\alpha(m)\sigma)/\sigma\sigma'$ to get 
$n/\sigma+\{\alpha(m)/\sigma'\}=(n\sigma'+\{\alpha(m)\sigma\}/\sigma\sigma'$
\end{Proof}



\begin{Lemma}
\textcolor[rgb]{1,0,0}{(We start off with a short exact sequence and localize its modules by  a set $\Sigma$. We want to see whether it still is a short exact sequence)
The localization of the modules (?):
$\alpha: M\rightarrow N$
$\Sigma^{-1}\alpha: \Sigma^{-1}M\rightarrow \Sigma^{-1}N
m/\sigma\mapsto \alpha(m)/\sigma$
}
If 
$$0\rightarrow M' \stackrel{\psi'}{\rightarrow}M\stackrel{\psi}{\rightarrow}M''\rightarrow 0$$
 is a short exact sequence of $R$-modules and $\Sigma$ is a multiplicative set in $R$, then 
$$0\rightarrow \Sigma^{-1}M'\stackrel{\Sigma^{-1}\psi'}{\rightarrow}\Sigma^{-1}M\stackrel{\Sigma^{-1}\psi}{\rightarrow}\Sigma^{-1}M''\rightarrow 0 $$
i.e. $\Sigma^{-1}:MOD_R\rightarrow MOD_{\Sigma^{-1}R}$ is an exact functor.
\end{Lemma}

\begin{Lemma} There exists a natural isomorphism
$\OO: (\Sigma^{-1}R\otimes_R-)\Rightarrow \Sigma^{-1}:MOD_R\rightarrow MOD_{\Sigma^{-1}R}$

$\Sigma^{-1}R\otimes_R M\rightarrow \Sigma^{-1} M$
$(r/\sigma)\otimes m \mapsto (rm)/\sigma$
$(1/\sigma\otimes m\mapstoreverse(m/\sigma))$

with commuting square
$$
\xymatrix{
R\otimes_R M \ar[d]_{\zeta_M \cong} \ar[r]_{j\otimes id_M}& \Sigma^{-1}R\otimes_R M  \ar[d]_{\OO_M \cong} \\
M \ar[r]_{\stackrel{\sim}{\Phi}} & \Sigma^{-1}M 
}
$$
\end{Lemma}

\begin{Lemma}[Serre 1956]
$\Sigma^{-1} R$ is a flat $R$-module. \textcolor[rgb]{1,0,0}{What is a flat module?}

(he proved it for the not commmutative $R$.
\end{Lemma}
\begin{proof}
By Tutorial 13 Exercise 3(c), this means that if $\alpha: M \rightarrow N$ is injective, then so is $id_{\Sigma^{-1} R}\otimes \alpha: \Sigma^{-1} R \otimes_R M \rightarrow \Sigma^{-1} R \otimes_R N.$

$$
\xymatrix{
\Sigma^{-1}R\otimes_R M \ar[r]_{id_{\Sigma^{-1}R}\otimes \alpha} \ar[d]_{\OO_M \cong} & \Sigma^{-1}R\otimes_R N \ar[d]_{\OO_N \cong}\\
\Sigma^{-1}M \ar@{>->}[r]_{\Sigma^{-1}\alpha} &  \Sigma^{-1} N
}
$$
$Ker(\Sigma^{-1}\alpha)\stackrel{4.17}{=}\Sigma^{-1}Ker(\alpha)=\Sigma^{-1} 0=0$
\end{proof}

\subsection{Local properties}
As before when $\Sigma=R-p$ with $p$ a prime ideal of $R$, write $M_p$ for $\Sigma^{-1}M$.
Property $P$ of $R$ (respectively module $M$) is a local property if
$R$ (respectively $M$) has $P$ $\Leftrightarrow$ $\forall$ prime ideal $p$ the localization $R_p$ (respectively $M_p$) has $P$.

\begin{Lemma}

For a module $M$, being zero is a local property. More precisely, the following are equivalent:

\begin{enumerate}[i)]
	\item $M=0$.
	\item $\forall$ prime ideals $p$, $M_p=0$.
	\item $\forall$ maximal ideals $m$, $M_m=0$
\end{enumerate}

\end{Lemma}
\begin{Proof}
Obviously $i)\Rightarrow ii) \Rightarrow iii)$

$iii)\Rightarrow i):$
Let $x\in M-\{0\}$. Then $Ann(x):=\{r \in R | rx=0\}$ is a proper ideal of $R$, so by Zorn's Lemma lies in a maximal ideal $m$.
Then $M_m=0$ implies $x/1 =0\in M_m$,so by (4.16) there exists $\nu R-m$ with $\nu x =0$, so $\mu\in (R-m)\cap Ann(x)$. CONTRADICTION!!
\end{Proof}

\begin{Proposition}
Let $\alpha: M\rightarrow N$ be an $R$-homomorphism. Write $\alpha_p=(R-p)^{-1}\alpha$.

\begin{enumerate}[a)]
	\item $\alpha$ is mono $\Leftrightarrow$ $\forall$ prime ideals $p$ $\alpha_p$ is mono $\Leftrightarrow$ $\forall$ maximal ideals $m$ $\alpha_m$ is mono.
	\item As a) but with mono replaced by epi, iso or zero
	\item $L\stackrel{\beta}{\rightarrow}M\stackrel{\alpha}{\rightarrow}N$ is exact iff exact $\forall$ prime ideals $p$ iff exact $\forall$ maximal ideals $m$
\end{enumerate}

\end{Proposition}
\begin{Proof}
a) $\alpha$ mono $\Leftrightarrow$ $Ker(\alpha)=0$ $\Leftrightarrow$ $Ker(\alpha)_p$ $\forall p$ $\Leftrightarrow$ $Ker(\alpha_p)=0$ $\forall$ prime ideals $p$ $\Leftrightarrow$ $\alpha_p$ mono $\forall$ prime ideals $p$

b) is similar.

c) Use the fact that this sequence is exact iff $\alpha\beta=0$ and the resulting map $Im(\beta)\hookrightarrow Ker (\alpha)$ is epi. So use b).
\end{Proof}